Ask question

Ask question

All categories

2 years ago

11

What is the speed of a wave that has a frequency of 125 Hz and a wavelength of 1.25 meters? Express your answer to the nearest w

hole number. 156.25 meters per second

2 answers:

qwelly [4]2 years ago

7 0

Answer:

The speed of the wave is 156.25 meter per second .

Explanation:

Use the formula for the relation between the speed of a wave and frequency.

$v=f \lambda$

Where f is frequency and $\lambda$ is wavelength .

f = 125 hz

$\lambda=1.25\ meters$

Putting all the values in the formula

$v=125\times 1.25$

$v=156.25\ meter\ per second$

Therefore the speed of the wave is 156.25 meter per second .

shtirl [24]2 years ago

3 0

156 is the answer. so 156.25 is almost the same thing, you just round. It's not hard. Thank you!!!

You might be interested in

What are 5 examples of balanced and unbalanced forces in your home?

ryzh [129]

Balance:
a book resting on a table
a car driving at 10 miles per hour in constant velocity
a cat sitting on a chair
a bulb that attach to the ceiling
your grandma sleeping on a bed
Unbalance:
your brother sprinting across the kitchen
a ball rolling at 5 m/s^2
your mom trying to run at 2 m/s^2 to spank you
you dropping your coffee mug on a floor
a cat jumping out of your bed
a tear from your eye falling through the floor
Hope this helps

8 0

3 years ago

An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at 12.8 m/s2 . At t1 the rocket e

Simora [160]

Answer:4.39 s

Explanation:

Given

initial velocity $u=0$

acceleration $a=12.8 m/s^2$

velocity acquired by sled in $t_1$ time

$v=0+at$

$v=12.8t_1$

distance traveled by sled in $t_1 s$

$v^2-u^2=2as$

$(12.8t_1)^2-0=2\times 12.8\times s_1$

$s_1=6.4\cdot t_1^2$

distance traveled in $t_2$ time with velocity $v=12.8t_1$

$s_2=v\times t_2$

$s_2=12.8\times t_1\times t_2$

$s_2=12.8\cdot t_1\cdot t_2$

$s_1+s_2=5.37\times 10^3$

$6.4t_1^2+12.8t_1t_2=5370$ ----1

$t_1+t_2=97.7 s$

$t_2=97.7-t_1$

substitute the value of $t_2$ in 1

we get

$6.4t_1^2-1250.56t_1+5370=0$

thus $t_1=\frac{1250.56-1194.33}{12.8}=4.39 s$

$t_1=4.39 s$

5 0

3 years ago

Differentiate can you help me plzzzzzzz

Schach [20]

Explanation:

Acceleration and the other term are the same physical phenomenon - just with opposite affects. Acceleration increases velocity, while the other term that begins with the letter "R" reduces.

***(Brainly's language filter is blocking the use of the second term in my answer.)***

Acceleration is defined as the rate of change of the velocity of an object per unit of time...

$a = \frac{dv}{dt}$

Acceleration would be a POSITIVE (+) quantity, while the r-word would be negative (-).

4 0

2 years ago

two charged particles repel each other with a force of 10N when they are 30cm apart if they are repel each other by a force of 4

KonstantinChe [14]

Answer:

The pattern between electrostatic force and distance can be further characterized as an inverse square relationship

5 0

1 year ago

A charge of 3.20 μC is held fixed at the origin. A second charge of 3.20 μC is released from rest at the position (1.25 m, 0.570

siniylev [52]

Answer:

a) $v = 7.137 m/s$ and b) $r = 1.832\ m$

Explanation:

We have to analyze this problem from the point of view of energy conservation. In this case there are two kind of energy, electric potential energy and kinetic energy. First, at $t = t_1$ there isn't relative movement between the two charges, so kinetic energy is zero and the total energy ( $E_T = E_p + E_k$ ) is just potential.

$E_T = E_p = \frac{k q_1 q_2}{r}$

where $k$ is the Coulomb constant, $q_1$ and $q_2$ are the two interacting charges, and $r$ is the distance between them.

Considering the fixed charge at (x1,y1) = (0,0) and the second one at (x2, y2) = (1.25, 0.57), the initial distance is

$r = \sqrt{(x_2-x_1)^2 + (y_2- y_1)^2} = \sqrt{(1.25)^2 + (0.57)^2} = 1.374\ m$ , then if

$k =8.987\times 10^9 N m^2/C^2$ and $q_1 = q_2 =3.2\times 10^{-6} C$ ,

$E_T = E_p = \frac{k q_1 q_2}{r} = \frac{8.987\times 10^9 * 3.2\times 10^{-6}*3.2\times 10^{-6}}{1.374} = 6.698 \times 10^{-2} Nm$ .

Now, at $t = t_2$ , $r \rightarrow \infty$ and $E_p \rightarrow 0$ . This means all the energy is kinetic

$E_T = E_k = \frac{1}{2}mv_f^2$ , so

$v_f = \sqrt{2E_T/m} =\sqrt{2 *6.698 \times 10^{-2} /0.00263} = 7.137 m/s$ (mass in Kg).

That would be the velocity when the second charge moves infinitely far from the origin.

For the second part we have that $v = v_f/2$ , so kinetic energy is

$E_k = \frac{1}{2}mv^2 = 1.674 \times 10^{-2} Nm$

and potential energy is

$E_p = E_T - E_k = 6.698 \times 10^{-2} - 1.674 \times 10^{-2} = 5.024 \times 10^{-2} Nm$

so the distance is

$r = \frac{k q_1 q_2}{E_p} = 1.832\ m$

6 0

2 years ago